6
$\begingroup$

NOT A DUPLICATE: Homeomorphism in the definition of a manifold for example is slightly different.

A manifold according to Wikipedia, a book by Spivak, and serveral other books has the following in common:

$\forall x\in M\exists n\in\mathbb{N}_{\ge 0}\exists U$ neighbourhood of $x$ such that $U$ is homeomorphic to (an open subset) of $\mathbb{R}^n$

I don't like the "neighbourhood" part as:

A set $U\subseteq M$ is a neighbourhood to $x$ if $\exists V$ open in $M$ with $[x\in V\wedge V\subseteq U]$

There is no requirement for $U$ to be open. It could be closed!

Problem:

Suppose that $U$ is homeomorphic to $\mathbb{R}^n$ by a function $f:U\rightarrow\mathbb{R}^n$, we know that $f$ is bijective and continuous by definition. This means it is surjective. Thus $f^{-1}(\mathbb{R}^n)=U$

By continuity of $f$ this means that $U$ is open in $M$

This is a contradiction, as $U$ need not be open.

I would be much happier if the definition was "there exists an open set containing $x$ that is homeomorphic to $\mathbb{R}^n$


The open subset of $\mathbb{R}^n$ part

The definition requires there exists a neighbourhood (not all neighbourhoods) homeomorphic to $\mathbb{R}^n$ is this the same as requiring the neighbourhood be homeomorphic to an open subset of $\mathbb{R}^n$?

Thoughts:

I understand that any open interval (in $\mathbb{R}$) is homeomorphic to all of $\mathbb{R}$ however the union of two distinct intervals is open but not homeomorphic to all of $\mathbb{R}$, using this sort of logic it suggests that:

I require (probably through the Hausdorff property) the ability to find a small enough connected (I suspect) open set. Then the two would be equivalent.

I could prove this if I assume the manifold has a countable topological basis (because then it is metricisable and I can use open balls) but I'd like to prove it for all manifolds.

$\endgroup$
  • 2
    $\begingroup$ You are correct that neighborhood is used inconsistently amongst multiple sources. Wikipedia, being written by the masses, is going to have inconsistencies like this. However, I will guess that a well crafted textbook, such as Spivak's, will use the word "neighborhood" in a way that avoids this inconsistency. $\endgroup$ – Lee Mosher Sep 13 '15 at 19:47
  • $\begingroup$ @LeeMosher to my knowledge it doesn't define it. Every time I've encountered "neighbourhood" it usually means "any subset" and they even use neighbourhood systems as a precursor to topologies. (Spivak's book on differential geometry assumes a lot of knowledge, which I do have! I just want to resolve the ambiguities now) - please don't forget there is a second part to the question $\endgroup$ – Alec Teal Sep 13 '15 at 19:48
  • $\begingroup$ 1) Any definition of topological manifold uses open neighborhoods; any ambiguity comes from the author's impreciseness rather than an ambiguity in the understood definition. 2) These are equivalent, Hausdorff or no. The reason for saying "open subset of $\Bbb R^n$" is partly consistency: when you're defining, say, complex manifolds, you have to use open subsets rather than all of $\Bbb C^n$. Using this as the convention for topological manifolds is a matter of taste. $\endgroup$ – user98602 Sep 13 '15 at 19:57
  • $\begingroup$ Look at the solution to exercise 57 here: math.fau.edu/kalies/mtg6316/exercise_solns51-100.pdf . It basically shows how just about everything about the two definitions of neighborhoods agree $\endgroup$ – Alan Sep 13 '15 at 19:57
  • $\begingroup$ @Alan that's a great link actually. I'm going to try and use that now to show that the definitions actually don't matter. Having said that I'm not having the easiest time and would love your take an on an answer (you seem to get I want to do it formally) $\endgroup$ – Alec Teal Sep 13 '15 at 21:12
2
$\begingroup$

The problem in the box you've labelled "Problem" is not a problem: the chart $f$ is required to be continuous on $U$, but not on $M$, so $f^{-1}(\mathbb{R}^n)$ could be closed in $M$ (although it will always be open too). The topological subspace $\{0, 1\} \times (0, 1)$ of $\mathbb{R}^2$ is a $1$-manifold in which every point has a neighbourhood that is homeomorphic with $\mathbb{R}$ and is both open and closed.

It doesn't make any essential difference in the definition of an $n$-manifold whether you require the domains of the charts to be neighbourhoods or open neighbourhoods or whether you require the ranges of the charts to be open subsets of $\mathbb{R}^n$ or the whole of $\mathbb{R}^n$: any point in an open subset of $\mathbb{R}^n$ has a neigbourhood that is homeomorphic with the whole of $\mathbb{R}^n$; this means that if $M$ is a topological space and $x$ is a point of $M$, then $x$ has a neighbourhood homeomorphic with $\mathbb{R}^n$ iff for some open $V \subseteq \mathbb{R}^n$, $x$ has a neighbourhood homeomorphic with $V$.

$\endgroup$
  • 1
    $\begingroup$ The last bit "neighbourhood that is homeomorphic to $\mathbb{R}^n$" is only half the proof. I am fine with that direction. I want to be convinced that these definitions are the same. $\endgroup$ – Alec Teal Sep 13 '15 at 20:17
  • 1
    $\begingroup$ Have you got a proof that the preimage of a chart will be both open and closed in the manifold? math.stackexchange.com/questions/872659/… this is more useful I think, which I believe shows (not read it all yet) that the definitions of neighbourhood are equivalent. $\endgroup$ – Alec Teal Sep 13 '15 at 20:18
  • 1
    $\begingroup$ Last comment, we're not talking about an $n$-manifold. I really would love a proof of what you've said (and the second half of what you address in the second paragraph) $\endgroup$ – Alec Teal Sep 13 '15 at 20:20
  • 2
    $\begingroup$ @AlecTeal: Open balls in Euclidean space are homeomorphic to $\Bbb R^n$. Such a $W$ exists by the definition of $V$ being open. $\endgroup$ – user98602 Sep 13 '15 at 20:54
  • 2
    $\begingroup$ I understood your question when I first read it. If you are happy that you can now answer it, good for you. $\endgroup$ – Rob Arthan Sep 13 '15 at 22:46
1
$\begingroup$

For a topological manifold, $M$ with topology $\mathcal{J}$

I've done half of the question. I have shown that:

$\forall x\in M\exists V\text{ neighbourhood of }x\exists n\in\mathbb{N}_{\ge0}[x\in V\wedge V\cong\mathbb{R}^n]$
$$\iff$$ $\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}[x\in U\wedge U\cong\mathbb{R}^n]$

Which is the first half done.

I will write the proof in later, I'm doing other things right now, but this is proved, no handwaving - formally proved!


The second part is that:

$\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}\exists V[V\in\mathcal{O}(\mathbb{R}^n)\wedge x\in U\wedge U\cong V]$ $$\iff$$ $\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}[x\in U\wedge U\cong\mathbb{R}^n]$

I've outlined this but not completed it.


With these two facts a manifold is a manifold regardless of:

  1. which notion of neighbourhood you use
  2. What you demand a neighbourhood is homeomorphic to

I will write the proof in later, I've done one on paper

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.